Passing char* to function ruins data?

Here's a project I just took off of the "waiting to die" pile. It is a super-simple money-managing program that reads data line-by-line from a txt file. Everything is Shibby except for when the name of the current account is read in. It is read just fine, however, when I pass it to the label(char*) function, it dies. I cannot display the char* in the label function without an error. Here's the name-reading portion of the code and the account label function.

Code:

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case NAME:
                        {
                                Reader.getline(buffer, 55);
                                if(Reader.eof())
                                {
                                        active = false;
                                        break;
                                }
                                cout << "DEBUG: Buffer at case <name> <ReadInAccounts()>:" << buffer << endl;//works
                                CurrentAccount.label(buffer);//Error at this call
                                                                SaveAccount(CurrentAccount);
                                read_state = MONEY;
                        }break;

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Code:

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int ACCNT::label(char *InputName)
{
        if(isnamed)
                name = InputName;
        else
        {
                name = InputName;
                isnamed = true;
        }
        //cout << "DEBUG: name passed into <label()>" << InputName << endl;//DEBUG... causes a crash
        //cout << "DEBUG: account name at <label()>: " << name << endl;//DEBUG... causes a crash
return 0;
}

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I'll note also, that if later in the program (after all of the data is read in) I try to access the name of the account, it is simply a blank string.

got whitespace?

Pardon?

What is "name"...is it another char* ?

If so, what do you think "name = InputName;" might be doing?

What's buffer declared as?

char* buffer = new char[100];

and name is also a char*. The funny thing is, the argument in label() cannot be displayed without an error, but buffer can.

You cant use = to copy a char*....use strcpy() or better std::string - the latter way will get rid of the new/delete & resource management too

Thank you, I'll do that- but my question is in that this causes an error:

Code:

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cout << "DEBUG: name passed into <label()>" << InputName << endl;

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Quote:

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Originally posted by CodeMonkey
Thank you, I'll do that- but my question is in that this causes an error:

Code:

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cout << "DEBUG: name passed into <label()>" << InputName << endl;

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On executing "name = InputName" you are getting 2 pointers to the same string stored on the heap. So immediately after, you can pass both pointers to cout and get a valid string. But later I'm guessing you are calling "delete [] buffer" - this tells the heap manager that the memory for the string is no longer needed and can be use for whatever. So later if you pass name to cout, you are passing a pointer to deallocated memory - which is undefined and could very likely crash your system